Past exam of the non-commutative rings course of the University of Oxford / 2023 / 3 / b / Solution

For the submodule $N\subseteq M$, define its subspace filtration and the quotient filtration on the quotient module $Q=M/N$ byThey are compatible filtrations of a module, since multiplication by $F_{i}R$ sends $N_j$ into $N_{i+j}$ and $Q_j$ into $Q_{i+j}$.

In each degree, inclusion and the quotient map induceThe first map is injective: if $n\in N_j$ belongs to $M_{j-1}$, then $n\in N\cap M_{j-1}=N_{j-1}$. The second map is surjective by the definition of $Q_j$.

For exactness in the middle, a class $m+M_{j-1}$ lies in $\ker {\pi }_j$ precisely when $m\in M_j$ and $m+N\in Q_{j-1}$. This means $m=m^{\prime }+n$ for $m^{\prime }\in M_{j-1}$ and $n\in N$. Since $m,m^{\prime }\in M_j$, one also has $n\in N_j$, andConversely, every class from $N_j$ maps to zero in $Q_j/Q_{j-1}$. Taking the direct sum over all degrees gives the short exact sequence of graded modulesThe maps respect the $\mathrm{gr}R$-action because they come from module homomorphisms compatible with the filtrations of a module. Consequentlyas $\mathrm{gr}R$-modules. This proves exactness of associated graded modules for induced filtrations.

Furthermore $A=\mathrm{gr}R$ is a Noetherian ring, and $\mathrm{gr}M$ is a finitely generated module. Its submodule $\mathrm{gr}N$ and its quotient module $\mathrm{gr}(M/N)$ are therefore finitely generated modules. Hence the induced subspace filtration and quotient filtration are both good filtrations for the order filtration.